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Dissociation constant of weak acid

I'm preparing for a chemistry lab relating to "Determination of the dissociation constant of a eak acid".

In part of the lab, I will b doing the following for measuring "Ka" for ethanoic acid using the indicator method:
1- Mix ethanoic acid (0.02M, 5cm3) and sodium ethanoate (0.02M, 5cm3).
2- Add bromocresol green (10 drops) and mix thoroughly.

And, in the introduction to the lab, I'm given this information:
Hln(aq) <-----> H+(aq) + ln-(aq)
ka(Hln) = [H+][ln-] / [Hln]
ka(Hln) for bromocresol green = 2 x 10^-5 mol dm-3

Provided that, how can I answer those questions:
1- What is the ratio of [ln-]/[Hln] in the CH3COOH/CH3COONa mixture?
2- Using this equation "ka(Hln) = [H+][ln-] / [Hln]", calculate [H+] in the CH3COOH/CH3COONa mixture.
3- What is the value of Ka for ethanoic acid?

Thanks.
Original post by SWEngineer
I'm preparing for a chemistry lab relating to "Determination of the dissociation constant of a eak acid".

In part of the lab, I will b doing the following for measuring "Ka" for ethanoic acid using the indicator method:
1- Mix ethanoic acid (0.02M, 5cm3) and sodium ethanoate (0.02M, 5cm3).
2- Add bromocresol green (10 drops) and mix thoroughly.

And, in the introduction to the lab, I'm given this information:
Hln(aq) <-----> H+(aq) + ln-(aq)
ka(Hln) = [H+][ln-] / [Hln]
ka(Hln) for bromocresol green = 2 x 10^-5 mol dm-3

Provided that, how can I answer those questions:
1- What is the ratio of [ln-]/[Hln] in the CH3COOH/CH3COONa mixture?
2- Using this equation "ka(Hln) = [H+][ln-] / [Hln]", calculate [H+] in the CH3COOH/CH3COONa mixture.
3- What is the value of Ka for ethanoic acid?

Thanks.


TBH I can't see how this lab works.

The mixture you are preparing is a buffer in which the acid and salt concentration are equal.

as ka = [H+][A-]/[HA]

and

[HA] = [A-]

then

Ka = [H+]

So you know the [H+] = 1.78 x 10-5

At the midpoint of an indicator:

[HIn] = [In-]

and

Ka(HIn) = [H+]

But as you cannot get other quantitative data by observing indicators, where you go from there IDK
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Original post by charco
TBH I can't see how this lab works.

The mixture you are preparing is a buffer in which the acid and salt concentration are equal.

as ka = [H+][A-]/[HA]

and

[HA] = [A-]

then

Ka = [H+]

So you know the [H+] = 1.78 x 10-5

At the midpoint of an indicator:

[HIn] = [In-]

and

Ka(HIn) = [H+]

But as you cannot get other quantitative data by observing indicators, where you go from there IDK


Thanks for your reply.

Let's take it step by step.

For now, is the ratio of [ln-]/[Hln] clear here? Can we say it is 1:1?
Original post by SWEngineer
Thanks for your reply.

Let's take it step by step.

For now, is the ratio of [ln-]/[Hln] clear here? Can we say it is 1:1?


If its the midpoint of the indicator, then yes as I stated earlier...
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Original post by charco
If its the midpoint of the indicator, then yes as I stated earlier...


Thanks really for your replies on this.

Can you just explain it further? What do you mean by midpoint? And, did you state earlier that it would be 1:1?
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Original post by charco
If its the midpoint of the indicator, then yes as I stated earlier...


Where did you get 1.78 x 10-5 mol dm-3 from?

In the data I have provided, I menationed that ka(Hln) for bromocresol green = 2 x 10-5 mol dm-3.

Does that mean that the ka of acetic acid is 2 x 10-5 mol dm-3?
Original post by SWEngineer
Where did you get 1.78 x 10-5 mol dm-3 from?

In the data I have provided, I menationed that ka(Hln) for bromocresol green = 2 x 10-5 mol dm-3.

Does that mean that the ka of acetic acid is 2 x 10-5 mol dm-3?


ethanoic acid = acetic acid

ka = 1.78 x 10-5

The midpoint of an indicator is half way between the colours of the acidic form and the basic form. At the midpoint the pH = ka of the indicator.

Your lab seems trivial. You are using an indicator with the ka value the same as the acid.
(edited 13 years ago)
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Original post by charco
ethanoic acid = acetic acid

ka = 1.78 x 10-5

The midpoint of an indicator is half way between the colours of the acidic form and the basic form. At the midpoint the pH = ka of the indicator.

Your lab seems trivial. You are using an indicator with the ka value the same as the acid.


So, based on this, let us come again to the first question:

1- What is the ratio of [ln-]/[Hln] in the CH3COOH/CH3COONa mixture?

Can I say, The ratio is 1:1? Do you agree with that?
Original post by SWEngineer
So, based on this, let us come again to the first question:

1- What is the ratio of [ln-]/[Hln] in the CH3COOH/CH3COONa mixture?

Can I say, The ratio is 1:1? Do you agree with that?


yes
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Original post by charco
yes


Very nice!

Now, if we come to the second question:

2- Using this equation ka(Hln) = [H+][ln-] / [Hln], calculate [H+] in the CH3COOH/CH3COONa mixture., provided that I was given that Ka(Hln) for bromocresol green = 2 x 10-5 mol dm-3

Can we say the following:

Since we are using an indicator of known dissociation constant, and [Hln] = [ln-], we can thus find [H+].
So, Ka = [H+]
And, thus, [H+] = 2 x 10-5 mol dm-3

Do you agree with that?
(edited 13 years ago)
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Original post by charco
ethanoic acid = acetic acid

ka = 1.78 x 10-5

The midpoint of an indicator is half way between the colours of the acidic form and the basic form. At the midpoint the pH = ka of the indicator.

Your lab seems trivial. You are using an indicator with the ka value the same as the acid.


When you say: pH = ka of the indicator., maybe you mean pH = -log Ka?
(edited 13 years ago)
[QUOTE="SWEngineer;30445326"]When you say: pH = ka of the indicator., maybe you mean pH = -log Ka?

When the indicator is at its mid point (see above for explanation) the pH of the solution is equal to the pka value of the indicator.

For example:
----------------
If the indicator midpoint is measured to be at pH = 4.7 then the pKa value of the indicator is 4.7

So the [H+] = 10-pKa

If the pH is 4.7 then the [H+] = 1.99 x 10-5 = ka of the indicator
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Original post by charco
yes


But, shouldn't we have the value of pH here in order to find the ratio?

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